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TS EAMCET · Chemistry · Chemical Bonding and Molecular Structure

Match the following:
The correct answer is \(\begin{array}{llllllll}A & B & C & D & A & B & C & D\end{array}\)

  1. A \(\begin{array}{llll}V & \text { I } & \text { III } & \text { II }\end{array}\)
  2. B \(\begin{array}{llll}V & \text { II } & \text { I } & \text { IV }\end{array}\)
  3. C \(\begin{array}{llll}\text { III } & V & \text { I } & \text { IV }\end{array}\)
  4. D \(\begin{array}{llll}\text { III } &IV & \text { I } & \text { II }\end{array}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\begin{array}{llll}\text { III } &IV & \text { I } & \text { II }\end{array}\)

Step-by-step Solution

Detailed explanation

(A) Trigonal planar \(\rightarrow\) (III) \(\mathrm{BF}_3\), because boron has three (3) valence electrons which are involved in formation of three sigma \((\sigma)\) bonds with \(3 \mathrm{~F}\)-atoms and has no lone pair \((l p)\) of electrons. (B) Tetrahedral \(\rightarrow\)…